Matrix inverse and bounds Suppose we have $\{M_n\}$ a sequence of matrices such that $a_n M_n^{-1}$ is bounded, with $a_n \to \infty$ as $n \to \infty$.
Is it true that $a_n(M_n + b_n)^{-1}$ is also bounded, if $b_n \to 0$ as $n \to \infty$?
I would think so, since $M_n^{-1}$ is bounded $M_n$ must be far from being singular, so the extra $b_n$ won't impact the inverse much.
However, I'm not sure how to prove it.
EDIT: Similar result for invertibility 
 A: I will assume that $b_n$ is a matrix.
If it is supposed to be a number instead, just replace my $b_n$ with $b_nI$ and everything works just the same.
I also assume that $a_n$ are positive real numbers.
Your first assumption that $a_nM_n^{-1}$ is bounded means that $a_n\|M_n^{-1}\|\leq C$ for some constant $C>0$.
To make $M_n+b_n$ invertible, it is easiest to use a Neumann series argument, which shows that $M_n+b_n$ is invertible if $\|b_n\|<\|M_n^{-1}\|^{-1}$.
Since $\|b_n\|\to0$ and $a_n\to\infty$, there is an index $N$ so that $n>N$ implies $\|b_n\|\leq\frac{a_n}{2C}$ and so $\|b_n\|\leq\frac12\|M_n^{-1}\|^{-1}$.
Then
$$
(M_n+b_n)^{-1}
=
M_n^{-1}-M_n^{-1}b_nM_n^{-1}+M_n^{-1}b_nM_n^{-1}b_nM_n^{-1}-\dots
$$
and the series converges.
In fact, we may estimate it as
$$
\|(M_n+b_n)^{-1}\|
=
\sum_{k=0}^\infty\|M_n^{-1}\|\|b_nM_n^{-1}\|^k
\leq
\sum_{k=0}^\infty Ca_n^{-1}2^{-k}
=
2Ca_n^{-1},
$$
so $\|a_n(M_n+b_n)^{-1}\|\leq 2C$ when $n>N$.
So yes, it is indeed asymptotically bounded.
But mind you that the inverse $(M_n+b_n)^{-1}$ might fail to exist for finitely many $n$.
